Length — Question 3

Question 3

The tip has not moved, but the ruler reading has

An object’s tip lies $h=2.0\ \mathrm{mm}$ above a ruler. An observer’s eye is at height $H=202.0\ \mathrm{mm}$ above the ruler and horizontal distance $d=20.0\ \mathrm{mm}$ to the right of the vertical through the tip. The observer reads $R=18.40\ \mathrm{mm}$ where the extended sightline meets the ruler.

For this geometry, similar triangles give $\lvert R-X\rvert/h=d/(H-h)$, where $X$ is the tip’s true projected coordinate. Treat the data as exact within the model.

1. Use the diagram to determine the sign of $R-X$. Find $X$.

2. A second eye position has the same height and horizontal distance but lies to the left of the tip. Predict its reading. Under precisely these symmetric conditions, what does the mean of the two readings give?

3. Describe two geometric ways to eliminate or reduce the parallax. Does averaging readings from arbitrary, unsymmetrical eye positions guarantee the same correction?

Question 3 - Solution

The tip has not moved, but the ruler reading has

(a) Determine the direction before applying the magnitude.

For an eye to the right, the ray extended downward through the elevated tip meets the ruler to the left of $X$. Thus $R-X<0$, with $$\lvert R-X\rvert=\frac{hd}{H-h} =\frac{(2.0)(20.0)}{202.0-2.0}\ \mathrm{mm} =0.20\ \mathrm{mm}.$$ Therefore $\boxed{X=R+0.20\ \mathrm{mm}=18.60\ \mathrm{mm}}$.

(b) Reverse the geometry, not the object.

$$\frac{R_{\mathrm{right}}+R_{\mathrm{left}}}2=\frac{(X-e)+(X+e)}2=X,\quad e=\frac{hd}{H-h}.$$

Moving the eye to the symmetric left position reverses the shift while leaving its magnitude unchanged. Its reading is $(18.60+0.20)\ \mathrm{mm}=\boxed{18.80\ \mathrm{mm}}$. The mean is $$\frac{18.40+18.80}{2}\ \mathrm{mm}=\boxed{18.60\ \mathrm{mm}}=X.$$

(c) Remove the geometric source.

Place the eye directly above the tip ($d=0$), or bring the tip and ruler scale into the same plane ($h=0$). Either makes the modeled shift zero. Increasing eye height while holding $h$ and $d$ fixed also reduces the shift, although it may make reading harder.

An arbitrary average is not guaranteed to cancel parallax. The shifts must be equal in magnitude and opposite in sign; different lateral distances or eye heights generally violate that condition. Repeated readings from one displaced eye position preserve the same bias.